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When I see this question $$\lim_{x\to\infty} \left(1+\frac1x\right)^x$$ I think the answer is $1.$ I arrive at this conclusion by plugging infinity in. The $1/x$ goes to $0,$ $1 + 0 = 1,$ and $1^∞$ is just $1.$

The answer to this question is $e$ though.

Plugging directly into this limit does not yield the correct value.

How can I tell in the future when it is safe to plug in infnity into the equation?

user
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The Calculus Calculator
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  • "plugging" infinity in doesn't work because this is an indeterminate form, recall that $A^B=e^{B\log A}$ and $B\log A$ is in the form $\infty \cdot 0$ – user Sep 26 '24 at 19:45
  • refer to https://math.stackexchange.com/q/136784/ – user Sep 26 '24 at 19:49
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    Why do you say $1^\infty=1?$ If $n\geq0$ is an integer and $y\geq 0,$ then then $(1+y)^n\geq 1+ny,$ by the binomial theorem. This means if $x=n, y=\frac1n,$ then $(1+1/x)^x\geq 1+n\cdot\frac1n=2.$ – Thomas Andrews Sep 26 '24 at 19:49
  • you can plug $\infty$ in all the times you don't face with an indeterminate form – user Sep 26 '24 at 19:52
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    $1^\infty$ is, in fact, what we call an "indeterminate form." So if $f(x)\to1$ and $g(x)\to\infty,$ we can't tell, absent information, what $f(x)^{g(x)}$ is. For example, if $f(x)=2^{1/x}$ and $g(x)=x,$ then $f(x)^{g(x)}=2$ for all $x.$ And if $g(x)=x^2,$ then $f(x)^{g(x)}\to\infty.$ – Thomas Andrews Sep 26 '24 at 19:52
  • Just work on the basis that it is never safe to plug infinity into the expression. – Adam Rubinson Sep 26 '24 at 20:21
  • @AdamRubinson For the OP: this is fine in many cases, for example when $\infty \cdot \infty$ or $\infty +\infty$ or $0/\infty$ or $0^{\infty}$ (with base positive). – user Sep 26 '24 at 20:38
  • Continuity.$\phantom{}$ – anomaly Sep 27 '24 at 00:13

2 Answers2

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This is one of the classic indeterminate forms of limit. The main warning sign of an indeterminate form, and hence a situation where "plugging in infinity" is particularly bad, is that you have two parts of the limit fighting to pull it in different directions so you can get different answers by plugging in infinity in different places.

With $y^x$, looking at what happens when $y \rightarrow 1$ and $x \rightarrow \infty$, if you try setting $y = 1$ then you get $1^x = 1 \rightarrow 1$ as $x \rightarrow \infty$, but if you instead start by taking $x \rightarrow \infty$ first then depending on the value of $y$ you have

$$y^\infty \rightarrow \begin{cases} \infty & y > 1 \\ 1 & y = 1 \\ 0 & 0 < y < 1 \end{cases}$$

which all don't depend on $y$, so they seemingly won't change when you take $y \rightarrow 1$. So even if we only look at the case for $y > 1$ (which matches the limit in your question), there's clearly tension between the base which is pulling the limit towards 1 and the exponent which is pulling it to an infinite value, meaning that any limit that looks like $1^\infty$ is indeterminate and you have to do extra work to determine the actual value - it's possible to have limits of this form that go to 1, and infinity, and every value in between.

ConMan
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    You seem to be assuming $y>1$ here. When $0\le y<1$, then $y^{\text{big}}\approx0$. – Mark S. Sep 27 '24 at 00:03
  • That's a fair point, which isn't specifically relevant to the limit in this question but is for the indeterminate form in general. I'll add a note to my answer. – ConMan Sep 27 '24 at 00:04
  • Thank you for your answer. I now understand that if I get one of the 7 intermediate forms from plugging in infinity then I can not solve it by plugging in infinity. – The Calculus Calculator Sep 27 '24 at 14:08
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First of all congratulations for asking a good question.

Let me talk about some basic things on limits.

Your question is $$\lim_{x\to \infty}\left(1+\frac{1}{x}\right)^{x}$$

I can also say that $$\lim_{x\to \infty}\left(1+\frac{1}{x}\right)^{x}=\lim_{x\to 0}\left(1+x\right)^{\frac{1}{x}}$$

Now let's think about the limits $$\lim_{x\to 0^{+}}\left(1+x\right)^{\frac{1}{x}}$$ and $$\lim_{x\to 0^{-}}\left(1+x\right)^{\frac{1}{x}}$$

Now if you clearly observe then you will positively say that the limits $$\lim_{x\to 0^{+}}\left(1+x\right)^{\frac{1}{x}}$$ and $$\lim_{x\to 0^{-}}\left(1+x\right)^{\frac{1}{x}}$$ are not equal.

Why ?

Because $$\lim_{x\to 0^{+}}\left(1+x\right)^{\frac{1}{x}}$$ will tend to $\infty$ and $$\lim_{x\to 0^{-}}\left(1+x\right)^{\frac{1}{x}}$$ will tend to $0$.

And everyone knows that $0$ & $\infty$ are not equal.

Therefore limits of the form of $1^{\infty}$ are undefined.

Hope you understood.

Best Wishes

IIT ROORKEE

IIT Roorkee
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